The lines x−y+z=1,3x−y−z+1=0  and x+4y−z=1,x+2y+z+1=0are

# The lines $x-y+z=1,3x-y-z+1=0\text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}x+4y-z=1,x+2y+z+1=0$are

1. A

Parallel

2. B

intersection

3. C

perpendicular and skew

4. D

non-perpendicular and skew

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### Solution:

The direction ratios of line of intersection of two planes $x-y+z=1,3x-y-z+1=0$are proportional to the vector $\left|\begin{array}{ccc}i& j& k\\ 1& -1& 1\\ 3& -1& 1\end{array}\right|=i\left(0\right)-j\left(-2\right)+k\left(2\right)$

The direction ratios of the line ${L}_{1}$is$〈0,1,1〉$

A point on the line $P\left(0,-1,0\right)$, by substituting  $z=0$solve the simultaneous equations.

The direction ratios of line of intersection of two planes $x+4y-z=1,x+2y+z+1=0$are proportional to the vector  $\left|\begin{array}{ccc}i& j& k\\ 1& 4& -1\\ 1& 2& 1\end{array}\right|=i\left(6\right)-j\left(2\right)+k\left(-2\right)$

The direction ratios of the line ${L}_{2}$ is $〈3,-1,-1〉$

A point on the line is $Q\left(-3,1,0\right)$, by substituting  $z=0$solve the simultaneous equations.

To check whether the lines lies in the same plane or not , find the value of $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|$

Hence, $\left|\begin{array}{ccc}{x}_{2}-{x}_{1}& {y}_{2}-{y}_{1}& {z}_{2}-{z}_{1}\\ {a}_{1}& {b}_{1}& {c}_{1}\\ {a}_{2}& {b}_{2}& {c}_{2}\end{array}\right|=\text{\hspace{0.17em}}\left|\begin{array}{ccc}0& 0& -2\\ 1& 2& 1\\ 3& -1& -1\end{array}\right|=14$, not equal to zero.

Therefore, the lines are skew lines and not perpendicular.  +91

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